C Halting Problem

题目链接

C Halting Problem

In computability theory, the halting problem is the problem of determining, from a description of an arbitrary computer program, whether the program will finish running (i.e., halt) or continue to run forever.

Alan Turing proved in 1936 that a general algorithm to solve the halting problem cannot exist, but DreamGrid, our beloved algorithm scientist, declares that he has just found a solution to the halting problem in a specific programming language -- the Dream Language!

Dream Language is a programming language consists of only 5 types of instructions. All these instructions will read from or write to a 8-bit register rrr, whose value is initially set to 0. We now present the 5 types of instructions in the following table. Note that we denote the current instruction as the iii-th instruction.

Instruction	Description
add vvv	Add vvv to the register rrr. As rrr is a 8-bit register, this instruction actually calculates (r+v)mod256(r + v) \mod 256(r+v)mod256 and stores the result into rrr, i.e. r←(r+v)mod256r \leftarrow (r + v) \mod 256r←(r+v)mod256. After that, go on to the (i+1)(i + 1)(i+1)-th instruction.
beq vvv kkk	If the value of rrr is equal to vvv, jump to the kkk-th instruction, otherwise go on to the (i+1)(i + 1)(i+1)-th instruction.
bne vvv kkk	If the value of rrr isn't equal to vvv, jump to the kkk-th instruction, otherwise go on to the (i+1)(i + 1)(i+1)-th instruction.
blt vvv kkk	If the value of rrr is strictly smaller than vvv, jump to the kkk-th instruction, otherwise go on to the (i+1)(i + 1)(i+1)-th instruction.
bgt vvv kkk	If the value of rrr is strictly larger than vvv, jump to the kkk-th instruction, otherwise go on to the (i+1)(i + 1)(i+1)-th instruction.

A Dream Language program consisting of nnn instructions will always start executing from the 1st instruction, and will only halt (that is to say, stop executing) when the program tries to go on to the (n+1)(n + 1)(n+1)-th instruction.

As DreamGrid's assistant, in order to help him win the Turing Award, you are asked to write a program to determine whether a given Dream Language program will eventually halt or not.

Input

There are multiple test cases. The first line of the input is an integer TTT, indicating the number of test cases. For each test case:

The first line contains an integer nnn (1≤n≤1041 \le n \le 10^41≤n≤10​4​​), indicating the number of instructions in the following Dream Language program.

For the following nnn lines, the iii-th line first contains a string sss (s∈{“add”,“beq”,“bne”,“blt”,“bgt”}s \in \{\text{``add''}, \text{``beq''}, \text{``bne''}, \text{``blt''}, \text{``bgt''}\}s∈{“add”,“beq”,“bne”,“blt”,“bgt”}), indicating the type of the iii-th instruction of the program.

• If sss equals to "add", an integer vvv follows (0≤v≤2550 \le v \le 2550≤v≤255), indicating the value added to the register;
• Otherwise, two integers vvv and kkk follow (0≤v≤2550 \le v \le 2550≤v≤255, 1≤k≤n1 \le k \le n1≤k≤n), indicating the condition value and the destination of the jump.

It's guaranteed that the sum of nnn of all test cases will not exceed 10510^510​5​​.

Output

For each test case output one line. If the program will eventually halt, output "Yes" (without quotes); If the program will continue to run forever, output "No" (without quotes).

Sample Input

4
2
add 1
blt 5 1
3
add 252
add 1
bgt 252 2
2
add 2
bne 7 1
3
add 1
bne 252 1
beq 252 1

Sample Output

Yes
Yes
No
No

Hint

For the second sample test case, note that rrr is a 8-bit register, so after four "add 1" instructions the value of rrr will change from 252 to 0, and the program will halt.

For the third sample test case, it's easy to discover that the value of rrr will always be even, so it's impossible for the value of rrr to be equal to 7, and the program will run forever.

思路:

代码:

#include<iostream>
#include<cstdio>
#include<cstring>
#define in(x) scanf("%d",&x)
#define out(x) printf("%d",x)
#define rep(i,a,b) for(i=a;i<=b;i++)
using namespace std;
const int maxn=10000+5;
int k[maxn],lucky[maxn];
int i,j,sum,r,flag;
const int mod=256;
struct node{
	string str;
	int v,k;
};
int main()
{
	string str;
	int t,a,v,k;
	in(t);
	while(t--){
		node pp[maxn];
		memset(lucky,0,sizeof(lucky));
		in(a);
		sum=r=0;
		rep(i,1,a){//模拟大法 
			cin>>pp[i].str>>pp[i].v;
			if(pp[i].str!="add"){
				cin>>pp[i].k;
			}
		}
		flag=0;
		i=1;
		while(1){//“add”,“beq”,“bne”,“blt”,“bgt”
		    lucky[i]++;	
			if(pp[i].str=="add"){
				r=(r+pp[i].v%mod)%mod;
				i++;
			}
			else if(pp[i].str=="beq"){
				if(pp[i].v==r) i=pp[i].k;
				else i++;
			}
			else if(pp[i].str=="blt"){
				 if(pp[i].v>r) i=pp[i].k;
				 else i++;
			}
			else if(pp[i].str=="bne"){
				if(pp[i].v!=r) i=pp[i].k;
				else i++;
			}
			else if(pp[i].str=="bgt"){
				if(pp[i].v<r) i=pp[i].k;
				else i++;
			}
			if(i>a) 
			{
				flag=1; 
			    break; 
			}   
			if(lucky[i]>=256) break;
		}
		if(flag==1) printf("Yes\n");
		else printf("No\n");
	}
	return 0;
}